Prove that if a body is thrown vertically upwards,then the time of ascent is equal to the time of descent.

(N/A) For a body thrown vertically upwards with initial velocity $u$,the final velocity at the maximum height is $v = 0$.
Let $t_{1}$ be the time of ascent. Using the first equation of motion $v = u + at$,where $a = -g$:
$0 = u - gt_{1}$
$\Rightarrow t_{1} = \frac{u}{g} ......(1)$
For the return journey (descent),the body starts from rest at the maximum height,so initial velocity $u' = 0$. The final velocity when it reaches the ground is $v' = u$. Let $t_{2}$ be the time of descent. Using $v' = u' + gt_{2}$:
$u = 0 + gt_{2}$
$\Rightarrow t_{2} = \frac{u}{g} ......(2)$
Comparing equation $(1)$ and $(2)$,we get $t_{1} = t_{2}$.
Thus,the time of ascent is equal to the time of descent.